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On what intervals of t is the curve described by the given parametric equation concave up? Concave down?

x=t^2; y=t^(3) + 3t

The signage of the second order derivative( ie positive, negative or zero) would help to discover the nature of concavity for the curve at various intervals.

dy/dx =dy/dt * dt/dx = (3t^2 + 3)/ (2t)

(d^2 y)/(dx^2)= d/dx (dy/dx) = d/dt (dy/dx) * ( dt/dx)

= [2t*(6t)-2(3t^2 + 3)]/(4t^2) * 1/(2t)

= (6t^2 - 6)/(8t^3) = (3t^2 - 3)/(4t^3)

(Note that since dy/dx is actually a function in t, implicit differentiation must be employed when seeking the expression for (d^2 y)/(dx^2) )

When the curve is concave downwards, (d^2 y)/(dx^2)<0 =====> (3t^2 - 3)/(4t^3) < 0

[(t-1)(t+1)]/ (t^3) < 0

ie t < -1 or 0 < t < 1 (shown)

When the curve is concave downwards, (d^2 y)/(dx^2) =====> (3t^2 - 3)/(4t^3) > 0

[(t-1)(t+1)]/ (t^3) > 0

ie -1 < t < 0 or t > 1 (shown)

Hope this helps. Peace.

Calculus

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